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有链表1 2 3 4 5 6 7 8 ,3个一组反转结果:3 2 1 6 5 4 8 7
Node* KNodeReverse(Node* root,int k)
{
Node *cur,*pre,*next;//必须有三个临时指针,因为cur需要知道连接的对象,同时不丢失它的下个对象
cur = root;
pre = nullptr;
Node* revRoot = new Node(0);//新建一个节点作为起点,返回前再delete
Node* first;
Node* connect = revRoot;
while(cur)
{
int n = k;
first = cur;
while(n-- && cur)
{
next = cur->pNext;
cur->pNext = pre;
pre = cur;
cur = next;
}
connect->pNext = pre;
connect = first;
}
Node *tempt = revRoot;
revRoot = revRoot->pNext;
delete tempt;
return revRoot;
}